Polynomials 2-Mark Questions Class 10: CBSE PYQs with Answers

Sum = 6 + (−3) = 3; product = 6(−3) = −18.

One suitable polynomial is x² − (sum)x + product.

One suitable polynomial: x² − 3x − 18

x² + 4x − 12 = x² + 6x − 2x − 12.

= x(x + 6) − 2(x + 6) = (x + 6)(x − 2).

Set each factor equal to 0.

Zeroes: −6 and 2

3p(x) = 3x² + 4x − 4.

= 3x² + 6x − 2x − 4 = (3x − 2)(x + 2).

Multiplying by the non-zero constant 3 does not change the zeroes.

3x − 2 = 0 or x + 2 = 0.

Zeroes: $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ and −2

x² + 7x + 10 = (x + 2)(x + 5).

The zeroes are α = −2 and β = −5.

α + β = −7 = −b/a, with a = 1, b = 7 and c = 10.

αβ = 10 = c/a.

Zeroes: −2 and −5; both relationships verified

x² − 25 = (x − 5)(x + 5).

The zeroes are α = 5 and β = −5.

Here a = 1, b = 0 and c = −25.

α + β = 0 = −b/a; αβ = −25 = c/a.

Zeroes: 5 and −5; both relationships verified

One suitable polynomial is x² − $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$x − $\frac{\text{(}\text{1}\text{)/(}}{\text{9}\text{)}}$.

Multiply by 18: 18x² − 9x − 2 = (6x + 1)(3x − 2).

6x + 1 = 0 or 3x − 2 = 0.

One suitable polynomial: 18x² − 9x − 2; zeroes: $\frac{\text{(}\text{−1}\text{)/(}}{\text{6}\text{)}}$ and $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$

α + β = 6 and αβ = 7.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{α²}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β²}\text{)}}$ = $\frac{\text{(}\text{(α + β)² − 2αβ}\text{)/(}}{\text{(αβ)²}\text{)}}$.

4($\frac{\text{(}\text{1}\text{)/(}}{\text{α²}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β²}\text{)}}$) = 4 × $\frac{\text{(}\text{36 − 14}\text{)/(}}{\text{49}\text{)}}$.

4($\frac{\text{(}\text{1}\text{)/(}}{\text{α²}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β²}\text{)}}$) = $\frac{\text{(}\text{88}\text{)/(}}{\text{49}\text{)}}$

15x² − 19x + 6 = 15x² − 10x − 9x + 6.

= 5x(3x − 2) − 3(3x − 2) = (5x − 3)(3x − 2).

Set each factor equal to 0.

Zeroes: $\frac{\text{(}\text{3}\text{)/(}}{\text{5}\text{)}}$ and $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$

Since p(5) = 0, 25(k − 1) − 5(4k + 1) + 10 = 0.

25k − 25 − 20k − 5 + 10 = 0.

5k − 20 = 0.

k = 4

One suitable polynomial is x² − x − 12.

x² − x − 12 = (x − 4)(x + 3).

Set each factor equal to 0.

One suitable polynomial: x² − x − 12; zeroes: 4 and −3

Multiply by 3: 21y² − 11y − 2 = 21y² − 14y + 3y − 2.

= 7y(3y − 2) + (3y − 2) = (7y + 1)(3y − 2).

Set each factor equal to 0.

Zeroes: $\frac{\text{(}\text{−1}\text{)/(}}{\text{7}\text{)}}$ and $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$

Other zero = 42 − 15 = 27.

Product = 15 × 27 = 405.

One suitable polynomial is x² − (sum)x + product.

One suitable polynomial: x² − 42x + 405

6p(x) = 6x² + x − 12 = 6x² + 9x − 8x − 12.

= 3x(2x + 3) − 4(2x + 3) = (3x − 4)(2x + 3).

Set each factor equal to 0.

Zeroes: $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$ and $\frac{\text{(}\text{−3}\text{)/(}}{\text{2}\text{)}}$

Other zero = $\frac{\text{(}\text{−20}\text{)/(}}{\text{4}\text{)}}$ = −5.

Sum = 4 + (−5) = −1.

One suitable polynomial is x² − (sum)x + product.

One suitable polynomial: x² + x − 20

4x² − 4x + 1 = (2x − 1)².

The zeroes are α = β = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$.

Here a = 4, b = −4 and c = 1.

α + β = 1 = −b/a; αβ = $\frac{\text{(}\text{1}\text{)/(}}{\text{4}\text{)}}$ = c/a.

Zeroes: $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ and $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$; both relationships verified

α + β = 4 and αβ = −12.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{α²}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β²}\text{)}}$ = $\frac{\text{(}\text{(α + β)² − 2αβ}\text{)/(}}{\text{(αβ)²}\text{)}}$.

= $\frac{\text{(}\text{16 + 24}\text{)/(}}{\text{144}\text{)}}$ = $\frac{\text{(}\text{40}\text{)/(}}{\text{144}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{α²}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β²}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{18}\text{)}}$

Write p(x) = 3x² + (2k + 1)x − (k + 5).

Sum = $\frac{\text{(}\text{−(2k + 1)}\text{)/(}}{\text{3}\text{)}}$; product = $\frac{\text{(}\text{−(k + 5)}\text{)/(}}{\text{3}\text{)}}$.

Given sum = half the product: $\frac{\text{(}\text{−(2k + 1)}\text{)/(}}{\text{3}\text{)}}$ = $\frac{\text{(}\text{−(k + 5)}\text{)/(}}{\text{6}\text{)}}$.

4k + 2 = k + 5, so 3k = 3.

k = 1

α + β = 8 and αβ = k.

α² + β² = (α + β)² − 2αβ.

40 = 8² − 2k, so 2k = 24.

k = 12

A zero makes the polynomial equal to 0.

(k − 1)(−3)² + k(−3) + 1 = 0.

9k − 9 − 3k + 1 = 0, so 6k = 8.

k = $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$

Reciprocal zeroes have product 1.

Product = $\frac{\text{(}\text{−(k − 2)}\text{)/(}}{\text{6}\text{)}}$ = 1.

−k + 2 = 6.

k = −4

Let the zeroes be 3t and 4t.

3t + 4t = −a, so a = −7t.

(3t)(4t) = b, so b = 12t².

12a² = 12(−7t)² = 49(12t²) = 49b.

Hence, 12a² = 49b

Here p ≠ 0. Let the zeroes be t and 6t.

7t = $\frac{\text{(}\text{14}\text{)/(}}{\text{p}\text{)}}$, so t = $\frac{\text{(}\text{2}\text{)/(}}{\text{p}\text{)}}$.

6t² = $\frac{\text{(}\text{8}\text{)/(}}{\text{p}\text{)}}$, so $\frac{\text{(}\text{24}\text{)/(}}{\text{p²}\text{)}}$ = $\frac{\text{(}\text{8}\text{)/(}}{\text{p}\text{)}}$.

Since p ≠ 0, 24 = 8p.

p = 3

The leading coefficient must be non-zero, so p ≠ −1.

Sum of zeroes = $\frac{\text{(}\text{−(2p + 3)}\text{)/(}}{\text{p + 1}\text{)}}$ = −1.

−2p − 3 = −p − 1, so p = −2.

At p = −2, the leading coefficient is −1 ≠ 0.

p = −2

α + β = 2 and αβ = −1.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{2α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{2β}\text{)}}$ + 3αβ = $\frac{\text{(}\text{α + β}\text{)/(}}{\text{2αβ}\text{)}}$ + 3αβ.

= $\frac{\text{(}\text{2}\text{)/(}}{\text{−2}\text{)}}$ + 3(−1) = −1 − 3.

$\frac{\text{(}\text{1}\text{)/(}}{\text{2α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{2β}\text{)}}$ + 3αβ = −4

Sum = 2 + (−3) = −1 = −(a + 1), so a + 1 = 1.

Product = 2 × (−3) = −6 = b.

a = 0 and b = −6

Since 3 is a zero, 2(3)² − 3(3) + p = 0.

18 − 9 + p = 0, so p = −9.

Sum of zeroes = $\frac{\text{(}\text{3}\text{)/(}}{\text{2}\text{)}}$, so the other zero = $\frac{\text{(}\text{3}\text{)/(}}{\text{2}\text{)}}$ − 3.

p = −9; other zero = $\frac{\text{(}\text{−3}\text{)/(}}{\text{2}\text{)}}$

α + β = 3 and αβ = −1.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ = $\frac{\text{(}\text{α + β}\text{)/(}}{\text{αβ}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{−1}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ = −3

Here p and q denote the zeroes; f(x) denotes the polynomial function.

p + q = $\frac{\text{(}\text{−1}\text{)/(}}{\text{6}\text{)}}$; pq = $\frac{\text{(}\text{−2}\text{)/(}}{\text{6}\text{)}}$ = $\frac{\text{(}\text{−1}\text{)/(}}{\text{3}\text{)}}$.

Since pq ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{p}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{q}\text{)}}$ − pq = $\frac{\text{(}\text{p + q}\text{)/(}}{\text{pq}\text{)}}$ − pq.

= $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{6}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{p}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{q}\text{)}}$ − pq = $\frac{\text{(}\text{5}\text{)/(}}{\text{6}\text{)}}$

Here p and q denote the zeroes; p(y) denotes the polynomial function.

p + q = $\frac{\text{(}\text{1}\text{)/(}}{\text{21}\text{)}}$; pq = $\frac{\text{(}\text{−2}\text{)/(}}{\text{21}\text{)}}$.

(1 − p)(1 − q) = 1 − (p + q) + pq.

= 1 − $\frac{\text{(}\text{1}\text{)/(}}{\text{21}\text{)}}$ − $\frac{\text{(}\text{2}\text{)/(}}{\text{21}\text{)}}$ = $\frac{\text{(}\text{18}\text{)/(}}{\text{21}\text{)}}$.

(1 − p)(1 − q) = $\frac{\text{(}\text{6}\text{)/(}}{\text{7}\text{)}}$

α + β = $\frac{\text{(}\text{5}\text{)/(}}{\text{6}\text{)}}$; αβ = $\frac{\text{(}\text{−3}\text{)/(}}{\text{6}\text{)}}$ = $\frac{\text{(}\text{−1}\text{)/(}}{\text{2}\text{)}}$.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ = $\frac{\text{(}\text{α + β}\text{)/(}}{\text{αβ}\text{)}}$.

= $\frac{\text{(}\text{5}\text{)/(}}{\text{6}\text{)}}$ ÷ $\frac{\text{(}\text{−1}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{−5}\text{)/(}}{\text{3}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ = $\frac{\text{(}\text{−5}\text{)/(}}{\text{3}\text{)}}$

Let the zeroes be t and 5t.

6t = $\frac{\text{(}\text{12}\text{)/(}}{\text{4}\text{)}}$ = 3, so t = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$.

5t² = $\frac{\text{(}\text{2k + 1}\text{)/(}}{\text{4}\text{)}}$, so $\frac{\text{(}\text{5}\text{)/(}}{\text{4}\text{)}}$ = $\frac{\text{(}\text{2k + 1}\text{)/(}}{\text{4}\text{)}}$.

2k + 1 = 5.

k = 2

α + β = $\frac{\text{(}\text{−2}\text{)/(}}{\text{3}\text{)}}$; αβ = $\frac{\text{(}\text{−1}\text{)/(}}{\text{9}\text{)}}$.

α² + β² = (α + β)² − 2αβ.

= $\frac{\text{(}\text{4}\text{)/(}}{\text{9}\text{)}}$ + $\frac{\text{(}\text{2}\text{)/(}}{\text{9}\text{)}}$ = $\frac{\text{(}\text{6}\text{)/(}}{\text{9}\text{)}}$.

α² + β² = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$

α + β = 3 and αβ = −5.

The new zeroes are 2α and 2β.

Their sum is 2(α + β) = 6; their product is 4αβ = −20.

One suitable polynomial: x² − 6x − 20

Here p ≠ 0, because the polynomial is quadratic.

α + β = $\frac{\text{(}\text{−q}\text{)/(}}{\text{p}\text{)}}$; αβ = $\frac{\text{(}\text{r}\text{)/(}}{\text{p}\text{)}}$.

α³β + β³α = αβ(α² + β²) = αβ[(α + β)² − 2αβ].

= $\frac{\text{(}\text{r}\text{)/(}}{\text{p}\text{)}}$($\frac{\text{(}\text{q²}\text{)/(}}{\text{p²}\text{)}}$ − $\frac{\text{(}\text{2r}\text{)/(}}{\text{p}\text{)}}$).

α³β + β³α = $\frac{\text{(}\text{r(q² − 2pr)}\text{)/(}}{\text{p³}\text{)}}$

α + β = $\frac{\text{(}\text{−14}\text{)/(}}{\text{8}\text{)}}$ = $\frac{\text{(}\text{−7}\text{)/(}}{\text{4}\text{)}}$; αβ = $\frac{\text{(}\text{3}\text{)/(}}{\text{8}\text{)}}$.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ = $\frac{\text{(}\text{α + β}\text{)/(}}{\text{αβ}\text{)}}$.

= $\frac{\text{(}\text{−7}\text{)/(}}{\text{4}\text{)}}$ ÷ $\frac{\text{(}\text{3}\text{)/(}}{\text{8}\text{)}}$ = $\frac{\text{(}\text{−14}\text{)/(}}{\text{3}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ = $\frac{\text{(}\text{−14}\text{)/(}}{\text{3}\text{)}}$

Sum = −9 + 6 = −3; product = −9 × 6 = −54.

One suitable polynomial is x² − (sum)x + product.

One suitable polynomial: x² + 3x − 54

α + β = $\frac{\text{(}\text{16}\text{)/(}}{\text{5}\text{)}}$; αβ = −2.

Since αβ ≠ 0, $\frac{\text{(}\text{α}\text{)/(}}{\text{β}\text{)}}$ + $\frac{\text{(}\text{β}\text{)/(}}{\text{α}\text{)}}$ = $\frac{\text{(}\text{(α + β)² − 2αβ}\text{)/(}}{\text{αβ}\text{)}}$.

= ($\frac{\text{(}\text{256}\text{)/(}}{\text{25}\text{)}}$ + 4) ÷ (−2) = $\frac{\text{(}\text{356}\text{)/(}}{\text{25}\text{)}}$ ÷ (−2).

$\frac{\text{(}\text{α}\text{)/(}}{\text{β}\text{)}}$ + $\frac{\text{(}\text{β}\text{)/(}}{\text{α}\text{)}}$ = $\frac{\text{(}\text{−178}\text{)/(}}{\text{25}\text{)}}$

α + β = 2; αβ = $\frac{\text{(}\text{−5}\text{)/(}}{\text{3}\text{)}}$.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{α²}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β²}\text{)}}$ = $\frac{\text{(}\text{(α + β)² − 2αβ}\text{)/(}}{\text{(αβ)²}\text{)}}$.

= (4 + $\frac{\text{(}\text{10}\text{)/(}}{\text{3}\text{)}}$) ÷ $\frac{\text{(}\text{25}\text{)/(}}{\text{9}\text{)}}$ = $\frac{\text{(}\text{22}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{9}\text{)/(}}{\text{25}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{α²}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β²}\text{)}}$ = $\frac{\text{(}\text{66}\text{)/(}}{\text{25}\text{)}}$

α + β = $\frac{\text{(}\text{6}\text{)/(}}{\text{5}\text{)}}$; αβ = $\frac{\text{(}\text{1}\text{)/(}}{\text{5}\text{)}}$.

α + β + αβ = $\frac{\text{(}\text{6}\text{)/(}}{\text{5}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{5}\text{)}}$.

α + β + αβ = $\frac{\text{(}\text{7}\text{)/(}}{\text{5}\text{)}}$

α + β = 5 and αβ = 4.

Since αβ ≠ 0, $\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ − 2αβ = $\frac{\text{(}\text{α + β}\text{)/(}}{\text{αβ}\text{)}}$ − 2αβ.

= $\frac{\text{(}\text{5}\text{)/(}}{\text{4}\text{)}}$ − 8.

$\frac{\text{(}\text{1}\text{)/(}}{\text{α}\text{)}}$ + $\frac{\text{(}\text{1}\text{)/(}}{\text{β}\text{)}}$ − 2αβ = $\frac{\text{(}\text{−27}\text{)/(}}{\text{4}\text{)}}$

α + β = $\frac{\text{(}\text{−7}\text{)/(}}{\text{2}\text{)}}$; αβ = $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$.

α² + β² + αβ = (α + β)² − αβ.

= $\frac{\text{(}\text{49}\text{)/(}}{\text{4}\text{)}}$ − $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{39}\text{)/(}}{\text{4}\text{)}}$.

α² + β² + αβ = $\frac{\text{(}\text{39}\text{)/(}}{\text{4}\text{)}}$

Reciprocal zeroes have product 1.

$\frac{\text{(}\text{−(p − 2)}\text{)/(}}{\text{6}\text{)}}$ = 1, so −p + 2 = 6.

p = −4

α + β = $\frac{\text{(}\text{15}\text{)/(}}{\text{25}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{5}\text{)}}$; αβ = $\frac{\text{(}\text{2}\text{)/(}}{\text{25}\text{)}}$.

Sum of new zeroes = $\frac{\text{(}\text{α + β}\text{)/(}}{\text{3αβ}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$.

Product of new zeroes = $\frac{\text{(}\text{1}\text{)/(}}{\text{9αβ}\text{)}}$ = $\frac{\text{(}\text{25}\text{)/(}}{\text{18}\text{)}}$.

One suitable polynomial is x² − $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$x + $\frac{\text{(}\text{25}\text{)/(}}{\text{18}\text{)}}$.

Multiply by 18 to obtain integer coefficients.

One suitable polynomial: 18x² − 45x + 25

Sum = (5 − 2√3) + (5 + 2√3) = 10.

Product = (5 − 2√3)(5 + 2√3) = 25 − 12 = 13.

One suitable polynomial is x² − (sum)x + product.

One suitable polynomial: x² − 10x + 13

Sum = 2 − $\frac{\text{(}\text{7}\text{)/(}}{\text{5}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{5}\text{)}}$.

Product = 2 × $\frac{\text{(}\text{−7}\text{)/(}}{\text{5}\text{)}}$ = $\frac{\text{(}\text{−14}\text{)/(}}{\text{5}\text{)}}$.

One suitable polynomial is x² − $\frac{\text{(}\text{3}\text{)/(}}{\text{5}\text{)}}$x − $\frac{\text{(}\text{14}\text{)/(}}{\text{5}\text{)}}$.

Multiply by 5 to obtain integer coefficients.

One suitable polynomial: 5x² − 3x − 14